Question: Is ${846993}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {846993}= &&{8}\cdot100000+ \\&&{4}\cdot10000+ \\&&{6}\cdot1000+ \\&&{9}\cdot100+ \\&&{9}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {846993}= &&{8}(99999+1)+ \\&&{4}(9999+1)+ \\&&{6}(999+1)+ \\&&{9}(99+1)+ \\&&{9}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {846993}= &&\gray{8\cdot99999}+ \\&&\gray{4\cdot9999}+ \\&&\gray{6\cdot999}+ \\&&\gray{9\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {8}+{4}+{6}+{9}+{9}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${846993}$ is divisible by $3$ if ${ 8}+{4}+{6}+{9}+{9}+{3}$ is divisible by $3$ Add the digits of ${846993}$ $ {8}+{4}+{6}+{9}+{9}+{3} = {39} $ If ${39}$ is divisible by $3$ , then ${846993}$ must also be divisible by $3$ Add the digits of ${39}$ $ {3}+{9} = \color{#9D38BD}{12} $ If $\color{#9D38BD}{12}$ is divisible by $3$ , then ${39}$ must also be divisible by $3$ $\color{#9D38BD}{12}$ is divisible by $3$, therefore ${846993}$ must also be divisible by $3$.